The correct option is C 11−√132
f(x)=x2+6x+c, c∈R
Since, f(x) is a polynomial of degree 2,
hence, f(f(x)) will be a polynomial of degree 4.
We know that, complex roots always occur in conjugate pair when co-efficients are real.
Since, f(f(x))=0 has three distinct real roots, so one root must be repeated.
Let roots of f(x)=0 be x=α,β
So, f(f(x))=0 has exactly three distinct real roots when α=−D4a and β>−D4a
⇒α=−(36−4c)4=c−9
Now, f(−D4a)=0
⇒f(c−9)=0
⇒(c−9)2+6(c−9)+c=0
⇒c2−11c+27=0
⇒c=11±√132
For f(x)=x2+6x+c=0, we have
α+β=−6
⇒β=3−c (∵α=c−9)
But, β>−D4a
⇒3−c>c−9
⇒2c<12
⇒c<6
∴c=11−√132